Calculus II
Practice Problems 9: Answers
In problems 1-5 find the radius of convergence of the series:
∞
1.
∑
2n
xn
n 1 !
n 1
Answer. We use the ratio test:
2n 1 n 1 !
n 2 ! 2n
2
n 2
0
Thus the radius of convergence is R ∞.
∞
2.
n
∑ 3 n xn
n 1
Answer. Again, the ratio test:
n 1 3n
3n 1 n
n 1 1
n 3
1
3
so R 3.
∞
3.
∑n n
1 n 2
n 0
x
3
n
Answer. We look at the ratios:
n 1 n n 1 3n
1
n
3
n n 1 n 2
n 1 1
n 2 3
1
3
2
1
n
so R 3.
∞ 4.
∑
n 1
2n ! n
x
n! 2
Answer. We look at the ratios:
2n 2 ! n! 2
n 1 ! 2 2n !
2n 2 2n 1 n 1 2
2
2 n
1
1 2
n
4
Thus R 1 4.
∞ 5.
∑
n 1
n 1 n 2 n 3 n
x
n!
Answer. Here we observe
that the general term is of the form a n p n n! xn , where p is a polynomial
of degree 3, Thus a n g n xn n 3 ! , where g is a ratio of polynomials of degree 3. Since lim g n exists, the numbers g n are bounded. Thus, by the comparison test (with the exponential series), this series
converges for all x; that is, R ∞.
∞ 6. Let f x ∑
n 0
n 2 n 1 n
x Find a formula for the function f .
n!
Answer. We have to look at the form of the series: is it obtained from a series we know using algebraic
operations, differentiation, or integration? The terms n 2 n 1 give a clue: these have been obtained by
two differentiations. So, we integrate twice to see what we have:
F x
∞
x
f t dt
0
G x
n
∞
x
F t dt
0
n 2 n
x
0 n!
∑
1
∑ n! xn
1
2
n 0
Now, we can factor out a x 2 , and obtain
G x
Then
F x
G x
Here’s another one to try:
x2 ex
∑
f x
∞ f x
1
n 0
2xex x2 ex
∞
∑ n! xn x2
F x
2ex 4xex x2 ex
n 2 n 1 x n
n 0
7. Find the Taylor series centered at the origin for the function F x Answer. We start with the series
x
0
dt
1 t4
∞
1
1 x
Substitute t 4
∑ xn n 0
for x, and then integrate term by term:
∞
1
1
F x
x
0
∑ t 4n t4 n 0
∞
dt
1
1
∑ 4n t4 n 0
1
t 4n 1
8. Find the Taylor series centered at the origin for the antiderivative (indefinite integral) of f x Answer. We start with the series
ex
Now, substitute
∞
xn
∑ n!
n 0
x2 for x and subtract 1:
e
x2
1
∞ ∑
n 0
1
nx
2n
n!
1
∞ ∑
n 1
1
nx
2n
n!
e
x2 x
1
Now, divide by x:
x2
e
∞ 1
∑
x
1
nx
2n 1
n!
n 1
Now integrate term by term:
∞
∑
f x
n 1
1 n 2n
x
2n n!
1 t2
dt
1 t2
x
9. Find the Taylor series centered at the origin for the function
0
Answer. We start with the series
∑ t 2n 1 t2
n 0
∞
t2
1 t2
so that
1 t2 ∞
1
Then
∑ t 2n
2
n 0
∞
1
1 t2
∞
∑ t 2n ∑ t 2n
n 0
2
n 0
Now, the second series in this sum is the same as the first, except for the first term. Thus
1 t2
1 t2
Now integrate term by term:
x
0
1 t2
dt
1 t2
∞
1 2 ∑ t 2n
∞
1
x2n
2n
1
1
x 2∑
n
10. Find the Taylor series centered at the origin for the function
Answer. We start with the series
1 x2
1
1 x2 2
∑ x2n n 0
Differentiate both sides;
2x
1 x2 1
∞
1
n 1
∞
2 ∑ 2nx2n 1
n 1
Now divide both sides by 2x, and change the index:
1
1 x2 ∞
2 ∑ nx2 n ∞ 1
n 1
∑
n 1 x2n
n 0
We remark that one could also start with the series for 1 1 x , differentiate it, and then substitute x 2 for x,
leading to, of course, the same result.
11. Find the Taylor expansion of x 3 centered at the point -1.
Answer.
We need to find
the successive derivatives of f x x 3 at x 1. We find: f 1 1 f 1 3 f 1 6 f 1 6, and all subsequence derivatives are zero. Thus the Taylor expansion is
x3
1 3 x 1 6
x 1
2!
2
6
x 1
3!
3
1 3 x 1 3 x 1
ex e 2
12. Find the Taylor series centered at the origin for the function cosh x Answer. We have
∞
ex
xn
∑ n!
n 0
∞ x
e
∑
nx
1
x 1
3
x
n
n!
n 0
2
When we add the two series, the odd terms cancel, and the even terms double. Thus
ex e so
coshx x
∞
2∑
x2n
2n !
ex e 2
n 0
∞
x
∑
n 0
13. Find the first 5 coefficients of the Maclaurin series for f x x2n
2n !
e x cos x.
Answer. We have to write down the Maclaurin series of e x and cos x up to the fifth term, and then multiply
just as we would polynomials, throwing away products which are of degree greater than 5;
ex cos x 1
x2
2
1 x
x4
24
x
14. Expand f x x2
2
x3
2
1
x3
6
x4
24
x5
120
x5 x2 x4 x3 x5
24 2
4
6 12
1 3 1 4 1 5
x x x x
3
6
30
x2
2
1
x4
24
x4
x5
24 120
1 x 3x 2 x9 in a Maclaurin series.
Answer. The expression of a polynomial as a sum of monomials is its expression as a Maclaurin series.